Acetone is treated with sodium amalgam and water gives

Options

A. `(CH_3)_2CHOH`
B. `CH_3CH_2CH_2OH`
C. `C_2H_5OH`
D. `CH_3CH_2COOH`

Explanation

To solve the question regarding the reaction of acetone with sodium amalgam and water, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reactants**:
- The reactant is acetone, which is the simplest ketone. Its chemical structure is CH3COCH3.

2. **Understand the Reaction Type**:
- The reaction involves the reduction of acetone. Sodium amalgam (Na/Hg) is a reducing agent that facilitates the reduction of ketones to alcohols.

3. **Determine the Product of the Reaction**:
- When acetone is reduced, the carbonyl group (C=O) in the ketone is converted to a hydroxyl group (C-OH), resulting in the formation of an alcohol.

4. **Identify the Type of Alcohol Formed**:
- The product formed from the reduction of acetone is a secondary alcohol because the carbon bearing the hydroxyl group is attached to two other carbon atoms.

5. **Write the Structure of the Product**:
- The structure of the product is CH3CHOHCH3, which is known as 2-propanol (or isopropyl alcohol).

6. **Check the Options Provided**:
- Based on the reduction of acetone, the correct product is 2-propanol. The other options (primary alcohols or carboxylic acids) do not correspond to the reduction of acetone.

7. **Conclusion**:
- Therefore, the final answer is that acetone treated with sodium amalgam and water gives 2-propanol.

### Final Answer:
Acetone treated with sodium amalgam and water gives **2-propanol**.

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